AMC12 2018 B

Answer (E): Let Chloe be $n$ years old today, so she is $n-1$ years older than Zoe. For integers $y \ge 0$, Chloe’s age will be a multiple of Zoe’s age $y$ years from now if and only if $$ \frac{n+y}{1+y}=1+\frac{n-1}{1+y} $$ is an integer, that is, $1+y$ is a divisor of $n-1$. Thus $n-1$ has exactly 9 positive integer divisors, so the prime factorization of $n-1$ has one of the two forms $p^2q^2$ or $p^8$. There are no two-digit integers of the form $p^8$, and the only one of the form $p^2q^2$ is $2^2 \cdot 3^2 = 36$. Therefore Chloe is 37 years old today, and Joey is 38. His age will be a multiple of Zoe’s age in $y$ years if and only if $1+y$ is a divisor of $38-1=37$. The nonnegative integer solutions for $y$ are 0 and 36, so the only other time Joey’s age will be a multiple of Zoe’s age will be when he is $38+36=74$ years old. The requested sum is $7+4=11$.